Find the general solution of the differential equation: $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $y = e^{3x} + C$ ##
Given differential equation is
$\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$
$\Rightarrow \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{e^x + \frac{1}{e^x}}$
$\Rightarrow \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{(e^{2x} + 1)} \cdot e^x$
$\Rightarrow \frac{dy}{dx} = 3e^{3x}$
$\Rightarrow dy = 3e^{3x} dx$
Integrating both sides, we get
$\int dy = 3 \int e^{3x} dx$
$\Rightarrow y = 3 \frac{e^{3x}}{3} + C$
$y = e^{3x} + C$